What is Compounding and Continuous Compounding of Interest?

What is simple interests and interest compounding? Full explanations with examples and easy to use online calculators to experiment with different values of the parameters involved in order to understand the different formulas related to interests compounding are presented. Problems on Compound interests with detailed solutions are also included in this site.

What is Percent Increase?

In order to understand compounding, you need to first understand the percentage increase of a quantity.

If P is a quantity that is increased by a percentage rate r, then the new quantity is P + r P

So if an amount P (principal) is invested at the annual rate r and is compounded annually, the total amount A at the end of t years is given by

A = P(1 + r) t

Example 2: $1000 is invested for 3 years, compounded every year, at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)

A = P(1 + r) t = 1000(1 + 0.03) 3 = $1092.73

You may use the calculator to input and experiment with more values for P, r and t and obtain the amount A. Use your own calculator and compare the results.

P = r = t = A = P(1 + r) t =

Interest Compounding n Times Per Year

How about compounding more that once a year? Let us say the interest is compounded twice a year (every 6 months) as follows:

Yearly rate is r; set a half yearly rate equal to r/2 and compound twice a year as follows:

t = 0 , A = P

At the end of the first 6 months of the year: A = P(1 + r/2)

At the end of the second 6 months of the same year: A = P(1 + r/2) + (r/2) P(1 + r/2) = P(1 + r/2) 2

We now extend the above to write:

So if an amount P (principal) is invested at the annual rate r and is compounded n times a year , the amount at the end of t years is given by

A = P(1 + r/n) n t

Example 3: $1000 is invested for 3 years, compounded twice a year (n = 2), at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)

A = P(1 + r/n) n t = 1000(1 + 0.03 / 2) 2×3 = $1093.44

You may use the calculator to input and experiment with more values for P, r, t and n and obtain the amount A. Use your own calculator and compare the results.

P = r = t = n =

A = P(1 + r/n) n t =

To understand the advantage of Compounding more than once a year, Keep P, r and t constant (The same amount invested at the rate r for t years) and increase n. What happens to A?

Continuous Compounding

So if an amount P (principal) is invested at the annual rate r and is compounded n times a year , the amount at the end of t years is given by (see above)

The question that one may ask is that what if we increase n indefinitely, which means increasing N indefinitely in our formula?

Use the calculator below to increase N ( 2 ,12, 1000, 10000...) and see how the value of the term (1 + 1 / N) N behaves?

N = (1 + 1/N) N =

As the number of compounding n increases, N also increases, the term (1 + 1 / N) N approaches a constant value which is called e and is approximately equal 2.718282... . More rigously, e is defined as the limit of (1 + 1/N) N as N approaches infinity.

Hence for continuous compounding (n very large) at the rate r and an initial amount P and after t years A is given by:

A = P e r t

Example 4: $1000 is invested for 3 years, compounded continuously, at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)

A = P e r t = 1000 e 0.03 × 3 = $1094.17

You may use the calculator to input and experiment with more values for P, r, and obtain the amount A. Use your own calculator and compare the results.

P = r = t = A =

Conclusion: compare the way the same amount of $1000 was compounded in the eaxamples 1,2 and 4 and make a conclusion as to which compounding earns more.